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I have an LED with the specs 3.6V, 20mA. The power supply I use is 5.1 volts.

To begin with, I used a 68 ohm resistor. When I measure parallel over the LED, I it measures 3.0 volts and when I connect the voltmeter in series series between the resistor and the LED, I get 31 mA.

According to this table, I should use a resistor at 75 ohms.

enter image description here

When I increase the resistance to 75 ohms it still results in 3.0 volts when I measure in parallel over the LED and the current in series between resistor and LED measures 27mA.

The voltage seems to be too low (3.0 v) and the current too high 27 mA. What resistor should I use to get it correct?

UPDATE I am using these two resistors in series for a total of 78 ohms:

enter image description here

enter image description here

I am going to have 20 LEDs connected to the same powersupply of 5.1 volt and 3 ampere.

I want the green LED light as bright as possible.

UPDATE 2 Here are some measurments on different resistors. If I use 120 ohm resistor I should be safe right?

  • 100 ohm: 2.98 v, 21.95 mA
  • 110 ohm: 2.96v, 20.42 mA
  • 120 ohm: 2.937v, 18.8 mA
  • 145 ohm: 2.89v, 15.73 mA
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    \$\begingroup\$ "... when I connect the voltmeter in series series between the resistor and the LED ..." You're then measuring current so you have probably switched to current measuring mode and your multimeter is now operating as an ammeter and not a voltmeter. \$\endgroup\$
    – Transistor
    Apr 4 at 19:43
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    \$\begingroup\$ If this is important, switching to 1% resistors will eliminate some of the variation (but not the LED forward voltage variation) for almost no extra cost. Switching to a constant current driver chip can maintain current uniformity across channels to a few percent. \$\endgroup\$
    – vir
    Apr 4 at 19:44
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    \$\begingroup\$ You probably do not want or need to run your LED at maximum brightness. A modern LED rated for 20mA can already blinding at 5mA. \$\endgroup\$
    – DKNguyen
    Apr 4 at 22:31
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    \$\begingroup\$ Be aware that the meter on the current range will have a significant resistance and will cause a voltage drop that will reduce the actual current compared to when the meter is not in the circuit. \$\endgroup\$ Apr 4 at 22:47
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    \$\begingroup\$ @KevinWhite, you're saying the DMM in current measurement will have a resistance high enough to significantly alter the current through a 68R resistor? I'd expect the DMM resistance to be far too tiny (sub-0.5ohm, with the probes about half of it) to even noticeably alter the current, with the ratio of the DMM resistance to the 68R. \$\endgroup\$
    – TonyM
    Apr 5 at 0:53

6 Answers 6

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  • Forward voltage drop for an LED can vary several hundred mV between different samples of the same part number
  • The LED forward current vs. forward voltage relationship is not linear
  • If you're using 5% resistors, 68 ohms at the high end of its tolerance band can overlap 75 ohms on the low end

If the LED forward voltage value is a maximum value and your 75 ohm resistor was low in the tolerance band, you will definitely see increased current. For example, if your actual forward voltage drop is only 3.1V and the resistor is actually 71 ohms, you will see more than 28mA.

Your options are to use a higher value resistor to bring current down to the desired value or use a current regulator/constant current driver.

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  • \$\begingroup\$ the goal is to have the current as close to 20 mA as possible?? (I want to keep the led as bright as possible) \$\endgroup\$
    – acroscene
    Apr 4 at 20:10
  • \$\begingroup\$ My guess is 3.6V is a maximum forward voltage number and the actual forward voltage is somewhere around 3.1V for green. I would try 110 or 120 ohms and see how that works. \$\endgroup\$
    – vir
    Apr 4 at 20:34
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The voltage across the LED is set by the colour of the LED and the characteristics of a diode. 3V is at the lower end of the Blue and Green LEDs. It can be as high as 3.6V. This voltage does not change much with current. There is some variation of course. Now that you know the actual voltage is 3.0V, then the resistor required is $$R=\frac{2.1V}{20\text{mA}}=105\Omega$$

The 3.6 Volt specification may be a maximum value. You should calculate the current with a minimum value for the voltage. If the LED voltage is higher then the current is lower.

On the other hand if you must have 20mA minimum, then calculate for the the higher voltage then a lower LED voltage will increase the current. Of course maximum ratings must also be observed.

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  • \$\begingroup\$ where does 2.1 come from? yes it is a green led \$\endgroup\$
    – acroscene
    Apr 4 at 19:38
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    \$\begingroup\$ @acroscene 5.1 -3=2.1 \$\endgroup\$
    – RussellH
    Apr 4 at 19:39
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    \$\begingroup\$ That is odd. Please post a link to the data sheet, and use the schematic editor in the question editor tool bar and post your actual circuit diagram. \$\endgroup\$
    – RussellH
    Apr 4 at 20:25
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    \$\begingroup\$ you are right. I did meassure wrong. Here is some meassurments on different resistors. If I use 120 ohm resistor I should be safe right? 100 ohm - 2.98 v, 21.95 mA || 110 ohm - 2.96v, 20.42 mA || 120 ohm - 2.937, 18.8 mA || 145 ohm - 2.89v, 15.73 mA \$\endgroup\$
    – acroscene
    Apr 4 at 21:33
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    \$\begingroup\$ @acroscene: That's better. Yes the 120 ohm resistor should work. \$\endgroup\$
    – RussellH
    Apr 4 at 22:56
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Most 5mm diameter through-hole LEDs have specs at a current of 20mA but their maximum allowed current is higher if you do not let them get too hot.

My LEDs are spec'd and recommended at 20mA and 40mA is their absolute maximum. But they will be a little brighter than 20mA if they operate at 30mA. LED

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  • \$\begingroup\$ could be that one. This is mine: electrokit.com/en/product/… \$\endgroup\$
    – acroscene
    Apr 4 at 21:48
  • \$\begingroup\$ @acroscene Yours is green. That one there is red. Not the same. \$\endgroup\$ Apr 5 at 3:48
  • \$\begingroup\$ @acroscene That site has no proper data for the LED. The 20mA could be the maximum current or just the rated current which but you don't know that. The voltage of 3.6V could be the maximum voltage at some high temperature at some current, but it might be lower, you don't know that either. It is useless to try to use LEDs with rather unknown parameters at some specific operating point. Just use 120 ohms and accept that they could be brighter and accept that some LEDs may be brighter than others. If you want to use LEDs better then you need to buy LEDs with known parameters. \$\endgroup\$
    – Justme
    Apr 5 at 5:45
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The 20 mA rating you give for the LED is likely an Absolute Maximum rating - exceeding that current may damage the LED.

LEDs are not fussy about current - they will work fine at much less than the Absolute Maximum rating, but do get dimmer at lower currents. If you are using the LED as an indicator light, you may find that it is bright enough for your needs at 10 mA or less. (I once had to reduce the current for a green LED to under 1 mA to get it dim enough for my application.)

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  • \$\begingroup\$ forgot to mention. I want it as bright as possible \$\endgroup\$
    – acroscene
    Apr 4 at 19:43
  • \$\begingroup\$ Most 5mm diameter through-hole LEDs have specs at a current of 20mA but their maximum allowed current is higher if you do not let them get too hot. \$\endgroup\$
    – Audioguru
    Apr 4 at 21:15
  • \$\begingroup\$ @acroscene - Producers like Osram are using 72% of the maximum current to drive their LEDs. In your case this would be 14.4mA of current over the LED. If you want to have more light, simply use an other LED with a bigger chip and a higher possible current. I have buy 1W LED for around 7-12 Cent for each. But they have to be soldered on a little aluminium cooling plate (6 Cent for each). \$\endgroup\$
    – MikroPower
    Apr 4 at 21:43
  • \$\begingroup\$ Do you mean I should consider 14.4mA as Maximum current instead of 20mA? \$\endgroup\$
    – acroscene
    Apr 4 at 22:02
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Just to summarize:

  1. "I am going to have 20 LEDs"
  2. "I want the green LED light as bright as possible"
  3. All using the "same power supply of 5.1 volt and 3 ampere"

Let's skip straight to the "as bright as possible" part. (There are so many other issues. But this is the starting point.)

Here's a datasheet for a similar jelly bean super bright green LED part. Let's look at some of the details.

enter image description here

From this, I can start with a few notes:

  1. There's about \$3.8\:\text{V}\$ at \$30\:\text{mA}\$, which works out to \$3.8\:\text{V}\cdot 30\:\text{mA}=114\:\text{mW}\$. I note that this is very close to their absolute maximum specification, which at the top mentions that ambient is \$25^\circ\$. This will yield the maximum operating temperature for the die (whatever that is.)
  2. The derating changes this to \$3.1\:\text{V}\$ at \$6\:\text{mA}\$ (and \$3.1\:\text{V}\cdot 6\:\text{mA}=18.6\:\text{mW}\$) when ambient is \$80^\circ\$. I can reasonably assume that this is when the maximum operating temperature for the die also occurs.
  3. So, assuming the limiting factor is the maximum die temperature, then the thermal resistance to air may be on the order of \$\frac{80^\circ-25^\circ}{114\:\text{mW}-18.6\:\text{mW}}\approx 580\:\frac{^\circ\text{C}}{\text{W}}\$, at a guess.

From this:

enter image description here

  1. Operating the LED at its absolute maximum current means that you must also operate at an ambient of \$25^\circ\$ and no more than that. And for all that trouble, you get about 20% more brightness by applying 63% more power into the device and without any room for an ambient temperature higher than \$25^\circ\$.
  2. Even operating the LED at \$20\:\text{mA}\$, the chart seems to be saying that you shouldn't operate at an ambient above about \$50^\circ\$.

The short and long of all this is that you do not want to operate such a device right at the very edge of its absolute maximum specifications. It's just not a good idea.

Obviously, there's more:

enter image description here

The typical specifications are saying that you might only get about 67% of the optical output from one LED and, by geometric implication anyway, perhaps 150% from a different one: even from the same part number and batch. This is a span of about 2.25:1. And that's when they are both operated at the same current.

Granted. Human brightness perception is logarithmic in nature. And a factor of 2.25:1 doesn't mean "twice as bright." But it likely means "noticeably different."

And that's brightness. Human color perception at this wavelength:

enter image description here

(from "Hue discrimination as a function of stimulus luminance" by Michael Siegel and Anne Siegel, 1972.)

...is about \$1\:\text{nm}\$. That's from an old paper dated in 1972, which follows work going all the way back to papers dating into the 1910's. Subsequent work still shows this same "W" effect over wavelength, though there's a lot more detail about all this, today. But the upshot is that humans can perceive wavelength shifts between two adjacent emitters with surprisingly good discernment. LEDs just cannot be produced (with any useful yield, anyway) that are tight enough that you won't notice the difference if you put them side by side.

So. Will you be doing that??? I've no idea. Only you know. But if so? You may notice annoying differences. This is part of why it's not easy to make LED displays where small details matter.

Finally, note that the voltage variation is also quite wide. From the specifications, it's not possible to know the entire span precisely. But you can certainly see there's quite a large difference from "typical" to "worst case." And the minimum isn't given.

A simple resistor as a current limiter is almost certainly not in the cards in this case. Not with such a low supply voltage rail and so much LED variation and 20 LEDs to handle. (But perhaps, if they are far apart from each other and your goals are low enough.)

You need to be a whole lot more specific about your application, my opinion. The variations are too wide, the margins too small, to be shooting from the hip.

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    \$\begingroup\$ The difference in wavelength / perceived color is not as big as a concern as you make it, I can tell from having bought and used millions of LEDs (though never at their absolute maximum ratings). Even without buying binned lots, if you buy them on a reel, they are usually not separable by perceived color. The rest of your answer nails it, though. \$\endgroup\$
    – markus-nm
    Apr 5 at 8:56
  • \$\begingroup\$ @markus-nm Thanks for the kind words. But I also worked for Siemens on LED display systems and LED manufacturing and binning. We'll just leave it there. \$\endgroup\$ Apr 5 at 9:07
  • \$\begingroup\$ Well, I've never used (or saw) Siemens-made or Siemens-tech LEDs, and I'm developing spectro/photometers, mostly with monochrome LEDs as source. I've been to the production line, seen the LEDs being tested before assembly. Sure you have 5-10 odd ones on a reel, but there is usually also something else off about these, not just the perceived color, and the reel is 4k units. \$\endgroup\$
    – markus-nm
    Apr 5 at 9:20
  • \$\begingroup\$ @markus-nm Yeah. I'm also using spectrophotometers and developed the software for the CIE hue and intensity analysis of same in order to bin them for their Penang factory. This both for individual discretes as well as display panels which must be color balanced to the D60 white point. I've been surprised just how good some people's color vision can be. I've also been surprised just how much of what color we see depends on what the surrounding colors also are. Very interesting. The resulting binning standards are quite high. \$\endgroup\$ Apr 5 at 18:33
  • \$\begingroup\$ @markus-nm I very much enjoyed reading some of Edwin Land's work from the late 1970's and early 1980's. Another discrimination paper is, "Measuring Wavelength Discrimination Threshold Along the Entire Visible Spectrum" by Ádám Krúdy and Károly Ladunga. This was interesting: "Wavelength Discrimination in the Presence of Added Chromatic Fields" by Pokorny and Smith. Among the many explanations, one I liked was "Human Wavelength Discrimination of Monochromatic Light Explained by Optimal Wavelength Decoding of Light of Unknown Intensity" by Zhaoping, Geisler, and May in 2011. \$\endgroup\$ Apr 5 at 18:44
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If I use 120 ohm resistor I should be safe right?

Yes, absolutely.

The forward voltage ratings on LEDs are ballpark only. LEDs don't make great absolute voltage references. If you keep them ovenized and running at a low current, they can be used as simple voltage references, but their absolute value will be "all over the place" - it's sensitive to manufacturing process variations.

Even Zener diodes are typically specified at +/-5% absolute accuracy. LEDs are worse than that usually.

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  • \$\begingroup\$ An important LED spec for brightness is its "viewing angle". There are some cheap LEDs in a tightly focusing case that makes then bright when shining directly at you but dim when you are a little off its focus axis. \$\endgroup\$
    – Audioguru
    Apr 6 at 1:04

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